uncool

That is not a definition. I am asking for a definition of what it means for the speed of light to be absolute. Again, only if you read it wrong. That is a part of what I want to show you, yes. So your objection is equivalent to the statement that if the (0, 0) element of Av (that is, the upper left element (which for this transformation matrix is [math] \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}[/math]) is greater than 1, then the (0, 0) element of (Av)-1 must be less than 1. Do you agree? Additionally, what is your transformation matrix? =Uncool-

Where did this number come from? Why is it important? This number never appears again. why do we care about it? This number never appears again except in the form of 144 itself. Why do the powers of 10 matter? Why not just say 12^2 = 144? Where does this number come from? Why not use 144 itself? Where does this number get used again? Do you really want to invite comment on that? =Uncool-

No, you did not, because you do not know what it means for the speed of light to be absolute. Show me precisely your definition of the absoluteness of the speed of light. What, precisely, does the statement that c is absolute mean? You are placing the parentheses wrong. Read as follows: [math](A^v)^{-1}[/math]. I wrote it with the brackets so that I could separate the superscript sections, as v and -1 should be considered completely separate, as v is an index, not an exponent. Again, what precisely does "absoluteness of light speed" mean in your "theory"? =Uncool-

That equation is a result of the absoluteness of the speed of light. What it means to say that the speed of light is absolute is that any observer in any frame will observe the speed of light to be c. You are saying that the speed of light is different in different frames, or in other words that all forms of relativity are false. That relativity is true comes entirely from experiment, so the absoluteness of the speed of light is an assumption. Anyone who has studied college level and even most high-school level math, yes. It denotes the {0, 0} element of the inverse matrix of Av, where I have already defined what Av is. Well, we can go through one by one. Av is the matrix I defined earlier. Then (Av)-1 is the inverse of Av. Finally, (Av)-10,0 denotes the 0,0 element of A - that is, the element in the upper left hand corner of A. Matrices are by no means obscure, and taking one element of a matrix is the entire point of having matrices in the first place. If you think they are obscure then you have not learned enough. And yes, I have an answer to your question; you are refusing to even think about the math behind that answer. That is because you have misunderstood what it means for the speed of light to be absolute. The speed of light being absolute means that for every observer in every inertial frame of reference, the speed of light is the same. In other words, the speed of light being absolute is the statement that if x^2 - (ct)^2 = 0, then x'^2 - (ct')^2 = 0. Which screen? You are talking about 2 screens here, not 1. Why must the two observers agree on ordering? =Uncool-

Actually, it really isn't sufficient. There are many ideas which can only be described mathematically. You are correct to say that this isn't one of them, but this is one of those ideas which is simpler to describe mathematically. You are trying to discuss physics. Matrices are not little-known to physicists - they are everywhere in physics. The basis was the assumption that the speed of light was absolute - that it was the same in every frame. It is possible to derive that the "cross-scale is absolute" from that assumption. Why would it be impossible for each of them to see a different sequence of frames? What precisely is wrong with that? And where does causality, which you invoked, come into it? =Uncool-

I have a good understanding of the idea. However, I cannot communicate it to you because of the inability to translate. Therefore, I am trying to write it in mathematical terms which translate more easily. That is the other advantage of math - it transcends language much more easily. If you cannot understand the math, then you have no place claiming to understand the physics. You say that the matrices are equivalent; then you should find it easy to translate from what I am saying to something you can easily understand. If you really want me to write it out without the matrices: Why? Why must you see an acceleration of time? Please give a precise reason why. =Uncool-

I am not trying to confuse the debate. I am trying to illustrate my point in a different way because you have not understood the other ways, and because this way can (to some people, including myself) be simpler. I am more at home with the matrix calculations here than I am at the word problem version of this that you are trying to use. The advantage of writing it mathematically is that it makes it abundantly clear what questions and answers are saying. That is the major problem with your assertions - they are not clear. By writing it mathematically, everything becomes clearer. The matrix is easier because the idea of an inverse happens to be important here. That is why I am trying to use it. =Uncool-

OK. So all of the theories (including your "Master Theory" and Einstein's special relativity) that you mention are linear theories, and can be described by transformation matrices. Do you agree that your statement about slowdowns is equivalent to saying that if the transformation matrix A for a velocity v is such that [math]A^v_{0,0} > 1[/math], then [math](A^{v^{-1}})_{0,0} < 1[/math]? On another note, what is your transformation matrix? =Uncool-

I have no idea what you are trying to say here. Again, this seems like gibberish. No, the math is based on an understanding of Einstein's theory. Do you understand matrices? If you do, then I can explain Einstein's theory. If you disagree with Einstein's theory, then the point where you disagree can be pointed out in the math just as well. Actually, you have not shown that. All you have done is claim it repeatedly. Please demonstrate that the principle of causality is violated in special relativity. =Uncool-

So you believe that light can travel at different speeds in different frames? For example, if you are moving at velocity v, then light will move v slower? =Uncool-

Alexander, I've been waiting for you to answer the question: If x^2 - (ct)^2 = 0, then a light ray sent from point 0 at time 0 will get to point x at time t, correct? =Uncool-

In relativity, the important point is that a spatial displacement in one frame becomes a displacement both in space and time in another frame. That is why time-ordering need not be consistent between frames. Actually, it doesn't. Relativity is entirely consistent with causality. Event A can only cause event B if event B is in the future light-cone of event A. As the light-cone is invariant, if it's true in one frame, it's true in every frame - which means that causality is entirely consistent with relativity. False. Do you know how matrices work? If so, I can explain the math via matrices. Now, will you finally answer the questions: If x^2 - (ct)^2 = 0, then a light ray sent from point 0 at time 0 will get to point x at time t, correct? =Uncool-

The two are moving away from each other. By definition the distance between the two is increasing. That is not the situation being discussed here at all. We are talking about two people who are moving away from each other at velocity v - which means that the distance between the two is increasing. Not quite. Time is no longer an object by itself. Again, not quite. What he says is that time can only be dilated in one way - that is, while in an inertial frame, time for another observer can only move more slowly than in one's own frame. Look up the twin paradox. =Uncool-

I mean that the distance between the two is becoming larger over time. Because space mixes with time in relativity, in another frame that means that time slows down. There is no contradiction between both thinking that the other has slowed down. If x^2 - (ct)^2 = 0, then a light ray sent from point 0 at time 0 will get to point x at time t, correct? =Uncool-

We're not talking about the transverse doppler effect - we're not talking about any doppler effect. I was answering your question. I have no idea what post you are responding to, because this has nothing to do with my post. Answer one at a time: If x^2 - (ct)^2 = 0, then a light ray sent from point 0 at time 0 will get to point x at time t, correct? =Uncool-

Because of the space that is growing between you two, slowed down. There is no contradiction. Read through the whole post to see why. Now will you answer my question? If x^2 - (ct)^2 = 0, then a light ray sent from point 0 at time 0 will get to point x at time t, correct? Then if we take a transform, the light ray must reach the point x' at time t' if (t', x') is the transformed version of (t, x), correct? In that case, since the speed of light is c in all frames, we then know that x'^2 - (ct')^2 = 0, correct? =Uncool-

So let's say that each one shows an image once a second from their own points of view. In other words: Observer A sees Observer B's ship moving at velocity v. A projects a screen at [math](0, 0, 0, 0), (1, 0, 0, 0), (2, 0, 0, 0)[/math], etc. relative to frame A. Correspondingly, Observer B sees Observer A's ship moving at velocity -v. B projects a screen at [math](0, 0, 0, 0), (1, 0, 0, 0), (2, 0, 0, 0)[/math], etc. relative to frame B. Then relative to frame A, A projects a screen at [math](0, 0, 0, 0), (1, 0, 0, 0), (2, 0, 0, 0)[/math]..., while B projects a screen at [math](0, 0, 0, 0), (\gamma, \gamma v, 0, 0), (2 \gamma, 2 \gamma v, 0, 0)[/math], ... Which says that the time-ordering (assuming low \gamma) should be the following: [math](0, 0, 0, 0)[/math]: Both project screen 0 [math](1, 0, 0, 0)[/math]: A projects screen 1 [math](\gamma, \gamma v, 0, 0)[/math]: B projects screen 1 [math](2, 0, 0, 0)[/math]: A projects screen 2 [math](2\gamma, 2\gamma v, 0, 0)[/math]: B projects screen 2 etc. However, relative to frame B, A projects a screen at [math](0, 0, 0, 0), (\gamma, -\gamma v, 0, 0), (2\gamma, -2\gamma v, 0, 0)[/math], ... while B projects a screen at [math](0, 0, 0, 0), (1, 0, 0, 0), (2, 0, 0, 0)[/math]... which gives the time-ordering of Which says that the time-ordering (assuming low \gamma) should be the following: [math](0, 0, 0, 0)[/math]: Both project screen 0 [math](1, 0, 0, 0)[/math]: B projects screen 1 [math](\gamma, -\gamma v, 0, 0)[/math]: A projects screen 1 [math](2, 0, 0, 0)[/math]: B projects screen 2 [math](2\gamma, -2\gamma v, 0, 0)[/math]: A projects screen 2 etc. The fact that space is mixing with time demonstrates why it is that the order can change in different frames. Now will you answer my question? If x^2 - (ct)^2 = 0, then a light ray sent from point 0 at time 0 will get to point x at time t, correct? Then if we take a transform, the light ray must reach the point x' at time t' if (t', x') is the transformed version of (t, x), correct? In that case, since the speed of light is c in all frames, we then know that x'^2 - (ct')^2 = 0, correct? =Uncool-

You are wrong; one seeing a slowdown does not imply that the other sees an acceleration, due to the mixing of space and time. However, that x^2 - (ct)^2 = 0 implies that x'^2 - (ct')^2 = 0 can be directly demonstrated from the assumption that c is constant: If x^2 - (ct)^2 = 0, then a light ray sent from point 0 at time 0 will get to point x at time t, correct? Then if we take a transform, the light ray must reach the point x' at time t' if (t', x') is the transformed version of (t, x), correct? In that case, since the speed of light is c in all frames, we then know that x'^2 - (ct')^2 = 0, correct? =Uncool-

That requires that if x^2 - (ct)^2 = 0, then x'^2 - (ct')^2 = 0. Do you agree with that? Edited to add: Einstein's theory actually doesn't make the claim that matter cannot move faster than light. The assumption only is that light can only move at c - everything is based off of that. Now, you have claimed that you agree that c is constant. That is enough to derive Einstein's equations, assuming you make the assumption about inertial frames. =Uncool-

So you are disputing the second assumption of relativity - that light must move at a velocity of c. That assumption actually does have reasoning; I was assuming that you agreed with it, and did not understand the reasoning that stemmed from that assumption. Do you agree that light is electromagnetic radiation? =Uncool-

What I just said is what Einstein wrote - the fact that transverse dimensions are not scaled is required by the theory, not a choice. =Uncool-

Read the entire section. He is specifically showing how to obtain the transform. It starts with two sections related only by a relative velocity v. Using the assumption that light move at speed c (which is required by Maxwell's equations), and the assumption that all inertial frames are equivalent, he then demonstrates the entire transform. Which choice are you referring to? That the transverse dimensions not be scaled? That is not a choice - it is required by the math. =Uncool-

Read section 3 - it details everything about the Lorentz transformation. =Uncool-

This is Einstein's original paper. Papers can serve as arguments in scientific debates - that is the entire point of papers in the first place. Could you write a specific version of what x', y', z', and t' are? In other words, could you give me your exact version of the equations I showed you? =Uncool-

I do see a reason. Have you ever read Einstein's original paper? http://www.fourmilab.ch/etexts/einstein/specrel/www/ Now, your post doesn't answer my question. I am looking for the actual transformation. What is the full transformation? In special relativity, the transformation is: [math] t' = \gamma (t - \frac{vx}{c^2})[/math] [math]x' = \gamma (x - vt)[/math] [math] y' = y[/math] [math] z' = z[/math] What is your version of the transformation? =Uncool-